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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A 1.5 Kg car moving on a flat , horizontal road negotiates a curve. If the radius of the curve is 35 m and the coefficient of static friction between the tires and dry pavement is 0.5, find the maximum speed the car can have and still make the turn successfully.

$\begin{array}{1 1} 12.2\;m/s \\ 3.5\;m/s \\ 4.8\;m/s\\ 13.1\;m/s\end{array} $

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Answer :13.1 m/s
In this case, the force that enables the car to remain in its circular path is the force of static friction.
$f_s =m \large\frac{v^2}{r}$
The maximum speed the car can have around the curve is the speed at which it is on the verge of skidding outward.
$v_{max} =\sqrt { \large\frac{f_{s, max}r}{m}}$
$\qquad= \sqrt {\large\frac{\mu_s m gr}{m}}$
$\qquad =\sqrt {\mu_s gr}$
$\qquad=\sqrt { (0.500)(9.80\;m/s^2)(35.0)}$
$\qquad= 13.1 \;m/s$
answered Aug 26, 2014 by meena.p

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